If frac{2x}{2x^2 + 5x + 2} frac{1}{x + 1},t | vedclass

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(C) Given the inequality: $\frac{2x}{2x^2 + 5x + 2} > \frac{1}{x + 1}$Factor the denominator: $\frac{2x}{(2x + 1)(x + 2)} > \frac{1}{x + 1}$Subtract $

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$-2 < x < -1$

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(C) Given the inequality: $\frac{2x}{2x^2 + 5x + 2} > \frac{1}{x + 1}$Factor the denominator: $\frac{2x}{(2x + 1)(x + 2)} > \frac{1}{x + 1}$Subtract $\frac{1}{x + 1}$ from both sides: $\frac{2x}{(2x + 1)(x + 2)} - \frac{1}{x + 1} > 0$Find a common denominator: $\frac{2x(x + 1) - (2x + 1)(x + 2)}{(2x + 1)(x + 2)(x + 1)} > 0$Simplify the numerator: $\frac{2x^2 + 2x - (2x^2 + 5x + 2)}{(2x + 1)(x + 2)(x + 1)} > 0$$\frac{-3x - 2}{(2x + 1)(x + 2)(x + 1)} > 0$Multiply by $-1$ and reverse the inequality: $\frac{3x + 2}{(2x + 1)(x + 2)(x + 1)} The critical points are $x = -2, -1, -\frac{2}{3}, -\frac{1}{2}$.Testing the intervals,the inequality holds for $x \in (-2, -1) \cup (-\frac{2}{3}, -\frac{1}{2})$.

If $\frac{2x}{2x^2 + 5x + 2} > \frac{1}{x + 1}$,then

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